Minuscule posets from neighbourly graph
sequences
N.J. Wildberger
School of Mathematics
UNSW
Sydney NSW 2052
Australia
February 7, 2001
1
Introduction
We construct minuscule posets, an interesting family of posets arising in Lie
theory, algebraic geometry and combinatorics, from sequences of vertices of
a graph with particular neighbourly properties.
We begin by associating to any sequence of vertices in a simple graph
X, here always assumed connected, a partially ordered set called a heap.
This terminology was introduced by Viennot ([11]) and used extensively by
Stembridge in the context of fully commutative elements of Coxeter groups
(see [8]), but our context is more general and graph-theoretic. The heap of a
sequence of vertices is that partially ordered set whose total linear orders correspond to all possible sequences obtained from the original one by switching
adjacent elements which are not neighbours in X. Furthermore sequences
which are equivalent under such interchanges (of adjacent elements which
are not neighbours in the graph) give rise to identical heaps.
A heap will be called neighbourly if the associated sequences have the
property that between any two successive occurrences of a vertex x there
occurs at least two occurrences of a neighbour of x.
Heaps arising from maximal neighbourly sequences which in fact have
exactly two neighbours between any two occurrences of a vertex x are classified. In our main result, we prove that any graph X having such a maximal
neighbourly heap which is in fact two-neighbourly must be one of the Dynkin
- Coxeter diagrams An , Dn , or E6 , E7 , and that the corresponding heaps are
exactly the minuscule posets defined and studied by Proctor in [4].
1
In the last section we briefly connect these interesting minuscule posets
(actually they are all distributive lattices) to Lie theory, algebraic geometry,
and combinatorics. This paper could be viewed as an elementary graph
theoretic approach to their study. We were led to these posets in our attempt
to construct Lie algebra representations directly from Dynkin diagrams, work
which is described in [12].
2
Neighbourly heaps for a graph
Let X be a simple graph. By an X-sequence we mean a sequence s =
(x1 , . . . , xn ) of vertices of X. If we transform s to s by switching xi and xi+1
for some i then there are three possibilities:
1) xi and xi+1 are neighbours in X – (an X-switch)
2) xi and xi+1 are distinct and not neighbours – (a free switch)
3) xi = xi+1 – (a redundant switch).
Any X−sequence s obtainable from s by free switches is defined to be
equivalent to s; we write s s and let [s] denote the equivalence class of
s, which we call an X-string. We refer to the xi in s = (x1 , . . . , xn ) as the
occurrences in s; as occurrences they are considered distinct even if as vertices
of X there may be repetitions. We partially order the occurrences xi in s by
declaring xi < xj if i < j and xi , xj are neighbours or identical vertices in X.
The resulting poset Ps is unchanged by free switches and so depends only on
the X-string [s]. We refer to Ps = P[s] as the X-heap of [s].
Proposition 2.1 The X-string [s] consists exactly of the total orderings of
P[s] consistent with the partial order.
Proof. Any sequence s obtained from s by free switches has the same heap
and so is an ordering of P[s] consistent with the partial ordering. Conversely
suppose s is an ordering of P consistent with the partial order. Let’s show
that we can free switch s to obtain s. Suppose by induction that s and s
agree up to to the kth term so that
s = (x1 , x2 , · · · , xk , xk+1 , · · · xn )
s = (x1 , x2 , · · · , xk , yk+1, · · · yn )
and that xk+1 = x. Clearly there is a first occurrence of x in yk+1 , · · · yn , and
if this first occurrence is preceded by a neighbour y = yj in X of x, then
since any two neighbours are necessarily related, we must have yj < xk+1 in
P . But this contradicts the fact that P is the heap of s, in which xk+1 occurs
before yj . 2
Example 1 Suppose X = An labelled as shown.
···
1
n−1 n
2
If we consider only X-sequences which are permutations of {1, . . . , n},
the associated heaps are ‘stock market graphs’ where each successive node
is either up or down from the previous. We get naturally a map from Sn to
the set of sequences {(η, . . . , ηn − 1) | η = ±1} = T. It is natural to ask for
the distribution of this map: how many permutations map to a given t ∈ T ?
When t is the zigzag sequence alternating plus and minus one, this is known
as André’s Problem, and the answer is given by Euler numbers, or Entringer
numbers. The general case has been recently solved by G. Szekeres.
Example 2 Suppose X = E6 labelled as shown
0
1
2
3
4
5
E6
The X-sequence s = (1, 2, 3, 0, 4, 5, 3, 2, 4, 3, 1, 0, 2, 3, 4, 5) has heap
5
4
3
2
1
0
3
2
4
3
5
0
4
3
2
F (E6 , 1)
1
For future reference, we refer to this particular heap as F (E6 , 1).
3
Definition An X-sequence s = (x1 , . . . , xn ) will be called neighbourly if
between any two consecutive occurrences of a vertex x there are at least two
occurrences of some neighbour or neighbours of x. This property is preserved
by free switches, so we also speak of neighbourly X-strings and X-heaps.
A neighbourly X-sequence s will be called maximal if F cannot be extended by the addition of a vertex x in any position to a larger neighbourly
X-sequence s , and similarly for X-strings and heaps. The neighbourly E6 heap of Example 2 is maximal.
A neighbourly X-string or X-heap will be called two-neighbourly if there
are exactly two occurrences of some neighbour or neighbours of x between any
two consecutive occurrences of any vertex x. The heap F (E6 , 1) of Example
2 is two-neighbourly.
Recall that a lattice is a poset such that for a, b ∈ L the least upper bound
a ∨ b and greatest lower bound a ∧ b exist uniquely. When these operations
satisfy the usual distributive laws, the lattice is called distributive. If P is any
poset, an ideal of P is a subset I such that x ∈ I, y ≤ x implies y ∈ I. Let
J(P ) denote the poset of all ideals of P ordered by inclusion. Then J(P ) is
always a distributive lattice, and any distributive lattice is of the form J(P )
for some poset P.
Proposition 2.2 If a graph X has a maximal neighbourly X-heap then X
is a tree.
Proof. If X is not a tree, consider the first occurrences of the elements of
some fixed cycle in X. The last occurrence in this set is necessarily preceded
by two neighbours, which contradicts maximality. Proposition 2.3 If F is a maximal neighbourly X-heap for some simple
graph X, then F is a lattice.
Proof. Let us suppose that P is a maximal neighbourly X-heap for some
graph X and that F = P[s] for some X-sequence s. The previous Proposition
shows that X must be a tree.
We first make a remark of general interest. Since the partial order on
F is generated by the relations xi < xj if i < j and xi , xj are neighbours
or identical in X, two occurrences xi and xj are related by xi < xj if and
only if there is a subsequence of s, xi = xi1 , xi2 , · · · , xik = xj such that
i1 < i2 < · · · ik (this is what we mean by a subsequence) and such that any
two successive elements in the subsequence are neighbours in X. That is, xij
and xij+1 are neighbours, for all j = 1, · · · , k − 1.
It can be useful to imagine that the vertices of X are lights which are
turned off and on in sequence according to s, so that the term xi in s means
that vertex x is lit up at time i. One is allowed to move from a vertex to a
4
neighbouring vertex precisely when that neighbouring vertex is lit. Then to
say that xi < yj is just to say that you can get from vertex x at time i to
vertex y at time j by a sequence of allowed moves.
Now suppose we have two occurences xi = x and xj = y in s with say
i < j. To say that there is a unique zk so that xi ≤ zk and yj ≤ zk is to say
that there is unique vertex on which two players A and B can meet at the
earliest possible time if they start at x and y at times i and j respectively.
Since X is a tree, if our two players want to meet as soon as possible
they will have to approach each other along the unique path which separates
them, say x = x0 , x1 , · · · , xk = y. This means that A will move to x1 at
the first opportunity, B will move to xk−1 at the first opportunity and so
on. If they can meet in this way it is clear that there is a unique vertex and
time when they will do so. Otherwise, they will reach a point when they are
unable to decrease the distance between them. Without loss of generality let
us assume this from the beginning. It means there is no occurence of x1 past
time i (and no occurence of xk−1 past time j).
But then by maximality there can be no occurrence of x2 past time i
either since then the previous occurence of x1 (which must exist) will be
followed by two occurences of its neighbours but not by another occurence of
itself, which is impossible. So after time i there is no occurence of x1 , x2 and
so on. But we are told that xk = y does occur after time i so our assumption
is impossible.
A similar argument shows that there is a unique occurence wl with wl ≤ xi
and wl ≤ yj . Recall the family of graphs Dn , n ≥ 4 and E7 and E8 labelled as shown
0
1
2
n−2 n−1
···
Dn
0
1
2
3
4
5
5
6
E7
0
1
2
3
4
5
6
E8
7
Theorem 2.1 Let X be a simple graph for which there exists a maximal
neighbourly X-heap F which is two-neighbourly. Then X is one of the graphs
An , n ≥ 1, Dn , n ≥ 4, E6 or E7 . There are exactly n such X-heaps for An ,
three for Dn , two for E6 and one for E7 .
The resulting X-heaps are precisely the set of minuscule posets defined
and studied in Proctor [4]. Let us illustrate what these minuscule posets look
like. Note that each is a coloured poset, where the colours are the vertices of
the top tree in each poset.
a) The case An . We label the minuscule An -heaps F (An , k) k = 1, . . . , n.
Hopefully the following example will make the general case clear.
For n = 5
5
4
4
3
5
3
3
4
2
2
1
4
1
3
3
1
F (A5 , 1)
2
2
4
2
3
F (A5 , 2)
F (A5 , 3)
2
1
1
3
2
2
4
3
3
5
4
4
F (A5 , 5)
F (A5 , 4)
5
b) The case Dn . The minuscule Dn -heaps are labelled F (Dn , 0), F (Dn , 1)
and F (Dn , n − 1). The following example for n = 5 should make the general
case clear.
6
5
1
0
4
2
0
2
1
3
2
1
3
4
0
1
3
2
1
2
2
4
3
3
0
2
2
3
1
F (D5 , 0)
4
F (D5 , 1)
F (D5 , 4)
The heaps F (Dn , 0) and F (Dn , 1) have the same triangular shape with
n(n − 1)/2 elements, while F (Dn , n − 1) consists of a square symmetrically
placed between two chains, and has 2(n − 1) elements.
c) The case E6 . There are two minuscule E6 -heaps labelled F (E6 , 1) and
F (E6 , 5). The heap F (E6 , 1) appeared in Example 2. The heap F (E6 , 5) has
the same shape, and is the inverse of F (E6 , 1).
1
2
3
0
4
3
2
5
4
1
3
2
0
3
4
5
F (E6 , 5)
d) The case E7 .
There is only one minuscule E7 heap labelled F (E7 , 6).
7
6
5
4
3
2
1
0
3
2
4
3
5
0
4
3
5
2
1
6
4
3
2
0
3
4
5
F (E7 , 6)
6
This lovely lattice, which we might call the swallow, is symmetric, spindle shaped, Sperner, Gaussian and enjoys other interesting combinatorial
properties (see [7],[9],[12]).
Note that in each case the graph X is an ideal of the minuscule X-heap
and that the minimal vertex appears in the label of that X-heap.
Proof of the Theorem. The proof will be broken down into several steps. We
will show that the assumption on s implies that X must be a tree with no
vertices of degree 4 or more and at most one vertex of degree 3. Then the
possibilities for this latter case will be analysed by reducing it to study of
triples of integers satisfying certain recursive properties. So let X and F be
given as in the theorem and let s be some X-sequence with heap F .
Lemma 2.1 X is a tree.
Proof. Let s0 be the sequence of initial occurrences of the vertices of X in s,
and P[s0] the associated heap, which clearly involves all vertices of X. Now
if X has a cycle then the largest element of this cycle appearing in P[s0 ] is
necessarily preceded by two neighbours, so contradicting maximality. Lemma 2.2 X cannot have a vertex of degree 4 or more.
8
Proof. Suppose X has a vertex e with neighbours a, b, c, d. Since each
occurs in s, e must occur at least twice.
Between the first and second occurrences of e we can have at most 2
occurrences of neighbours of e – that means, say, that c and d do not occur.
But then both c and d must occur before the first occurrence of e (if they
didn’t, we could add them, contradicting maximality) so we can add another
e to the front of the sequence which is impossible. Lemma 2.3 X cannot have two vertices of degree 3.
Proof. If X has at least two vertices of degree three then it has a subgraph
Y of the following form
(n + 1)
0
Dn
0
1
···
n
n+1
Consider the first occurrences in s of the vertices of the subgraph Y
and the associated heap PY . If the occurrences of the vertices 1 and n are
unrelated in PY then an easy argument shows that PY must have the following
form for some k, 1 < k < n.
k
···
···
n−1
2
n
1
0
0
n+1
(n + 1)
That means that the next occurrence of either 1 or n must precede the
next occurrence of 2 or n − 1, that then the next occurrence of 2 or n − 1
must precede the next occurrence of 3 or n − 2 etc. But that will imply that
the next occurrence of k is preceded by more than two of its neighbours, a
contradiction. On the other hand if say 1 < n in PY then again an easy argument shows
that the associated heap PY must have up to relabelling the following form.
9
0
1
0
2
···
n
n+1
(n + 1)
But then the next occurrence of n must precede the next occurrence of
n − 1, which must precede the next occurrence of n − 2 and so on down to
i, which is then preceded necessarily by three occurrences of neighbours of
itself since its first occurrence, again a contradiction. Now suppose that X has exactly one vertex, call it d, of degree 3, with
chains of length α, β, γ > 0 emanating from it, labelled a1 , a2 , . . . , aα , b1 , b2 ,
. . . , bβ and c1 , c2 , . . . , cγ as shown.
cγ
c2
c1
• ...
aα a2
a1
b1
d
... •
bβ
b2
We imagine weighting the vertices linearly as follows
d > c1 > c2 > · · · > cγ > b1 > b2 > . . . > bβ > a1 > a2 > · · · > aα
and make the convention that wherever possible lighter elements move forward by free switches in a sequence s (and so upwards in the reverse Hasse
diagram for P[s] ). In other words ai aj is replaced by aj ai if i > j and
|i − j| = 1, daj is replaced by aj d if j = 1 (and similarly with bsi , csi ) and bi aj
is replaced by aj bi , etc. The weighting above then induces a partial order on
elements of an X-string [s] so that there is a unique minimal X-sequence t
where no further free switches of the above type are possible.
Let us look in t at the successive occurrences of d and refer to the ith
interval of t as the segment following the ith d and before the (i + 1)st d (if it
occurs), for i = 1, · · · , r.
10
Lemma 2.4 For any i, 1 ≤ i ≤ r, there are non-negative integers αi , βi , γi
such that the ith interval has the form
a1 a2 . . . aαi b1 b2 . . . bβi c1 c2 . . . bγi .
Proof. Since all the aj can be freely switched with all the bj and all the cj
and the bj with the cj , the fact that the aj are lighter than the bj which are
lighter than the cj means that the ith interval will consist of a sequence of aj
followed by a sequence of bj followed by a sequence of cj with some of these
sequences possibly empty.
The first aj must be a1 , otherwise it would switch with d out of the ith
interval. The second aj must be a2 since it cannot be a1 and any other aj
would freely switch to the left out of the interval. Continuing, we must start
with a maximal sequence of aj of the form a1 a2 . . . aαi for some αi ≤ α. But
then the neighbourly condition ensures that no more αj are possible. Since
the bj and cj sequence are subject to the same analysis, the Lemma is proved.
Let us represent the sequence
a1 a2 . . . aαi
by the shorthand symbol aαi .
Proposition 2.4 If there are r intervals then t has the form
t = ···
d(1) aα1 bβ1 cγ1 d(2) aα2 bβ2 cγ2 d(3) · · ·
d(r) aαr bβr cγr ,
where d(k) is the k th occurrence of d and where the αi , βi , γi satisfy
1. for i = 1, . . . , r − 1 exactly one of αi , βi , γi is zero
2. for i = r exactly two of αi , βi , γi is zero
3. if αi > 0 then αi+1 = αi − 1 for i = 1, · · · , r − 1 (and similarly for βi
and γi )
4. if αi = 0 for i = 1, . . . , r − 1 then αi+1 > 0 (and similarly for βi and
γi).
Proof. If there are r intervals then let us show that t cannot end in d(r+1) .
If two of αr , βr , γr were non-zero, say αr and βr , and there was an (r + 1)st
occurrence of d, then by maximality another c1 could be added after this,
contradicting the assumption of r intervals. This also proves 2). Statement
1) is a consequence of the two neighbourliness of t.
11
Lets prove 3). Suppose αi > 0 for some i = 1, . . . , r−1. Then αi+1 ≥ αi is
impossible since the element aαi in the ith interval is then separated from the
aαi in the (i+1)st interval by a single neighbour, namely aαi −1 if αi > 1 or d if
αi = 1. Now if αi+1 < αi − 1 then there must be a following occurrence (after
the (i + 1)st interval) of aαi+1 +1 , since two neighbours of it have occurred.
But when it does occur next it does so with aαi+1 preceding it — meaning at
least 3 neighbours between occurrences.
To prove 4), note that if αi = 0 and αi+1 = 0 then 3 d’s will have occurred
between the previous a1 and the following a1 . Without loss of generality we may assume that α1 > 0, β1 > 0 and γ1 = 0.
This means there is necessarily by maximality an occurrence of c1 before the
first d.
Lemma 2.5 The portion of t before the first occurrence of d is
t = cγ cγ−1 , · · · c1 d(1) · · · .
Proof. We first note that no aj or bj may precede d(1) . Since c1 does occur
before d(1) , neither a1 of b1 can otherwise we could add another occurrence of d
to the beginning of the sequence. But then neither a2 or b2 can occur, because
otherwise we could add an a1 or b1 before it, contradicting the previous
statement. Continuing we obtain the claim.
To see that c1 is necessarily immediately to the left of d(1) , observe that
any cj , j > 2, is freely switched to the left of the c1 occurrence immediately preceding d(1) . If c2 occurs between this c1 and d(1) then since γ1 = 0
(assumption) there are three neighbours of c1 between it and the next occurrence of it after d(2) , which is impossible. Similarly the next previous cj
must be c2 , then c3 and so on. If as we proceed left from d(1) in t we find two
occurrences of cj then there must also be two occurrences of cj−1 , of cj−2 ,
and so on until two occurrences of c1 mean another d can be added to the
beginning, which is impossible. Thus t has the prescribed form. If we agree to write cγ cγ−1 , · · · c1 as c−γ then we see that t has the form
t = c−γ d(1) aα1 bβ1 d(2) aα2 bβ2 cγ2 · · · d(r) aαr bβr cγr ,
where we now analyse the possibilities for the sequence of triples
(0, 0, −γ), (α1 , β1 , 0), (α2 , β2 , γ2 ), · · · (αr , βr , γr ).
We know α1 , β1 , γ2 > 0. Since at least one of α2 , β2 , γ2 is zero, without loss
of generality we may assume that β2 = 0 so that β1 = 1 from 3) or 4) of
Proposition. The above sequence of triples is then of the form
(0, 0, −γ)(α1 , 1, 0)(α1 − 1, 0, γ2) · · · .
12
Lemma 2.6 β = 1.
Proof. If β > 0 consider first occurrence of b2 . It is then preceded by two b1 ’s,
so we may add b2 to the beginning of t contradicting the previous Lemma. Suppose now that r = 2. Then since 2 of α2 , β2 , γ2 are zero and γ2 we
know is not, we must have α2 = 0 so that α1 = 1. By maximality γ0 = γ2 = γ
and so the sequence of triples for t is
(0, 0, −γ), (1, 1, 0), (0, 0, γ).
This corresponds to X = Dn
0
1
2
n−2 n−1
···
Dn
and the sequence
t = (n − 1, n − 2, · · · 3, 2, 1, 0, 2, 3, · · · , n − 1).
In the case n = 5 the associated heap has the form
4
3
2
0
1
2
3
4
Suppose now that r > 2. Then exactly one of (α2 , β2 , γ2 ) = (α1 − 1, 0, γ2 )
is zero, so that (α3 , β3 , γ3) = (α1 − 2, 1, γ2 − 1). If r = 3 then both α1 − 2
and γ2 − 1 must be 0, giving α1 = 2, γ2 = 1 and the only possible maximal
form for the sequence of triples being
(0, 0, −1)(2, 1, 0)(1, 0, 1)(0, 1, 0).
This corresponds to X = D5 with sequence
t = (1 2 3 4 0 2 3 1 2 0)
13
and heap
1
2
0
3
2
4
1
3
2
0
F (D5 , 0)
If r > 3 then exactly one of (α3 , β3 , γ3 ) = (α1 − 2, 1, γ2 − 1) is zero.
We consider the 2 cases α1 = 2 and γ2 = 1 separately.
Case α1 = 2: If α1 = 2, γ2 > 1 then the triple sequence for t must have the
form
(0, 0, −γ)(2, 1, 0)(1, 0, γ2)(0, 1, γ2 − 1)(α4 , 0, γ2 − 2).
Now α4 must be 2, since α4 > 0 by Proposition and if α4 = 1 then the
next occurrence of a2 (which must occur) will have (at least) three neighbours
between it and the first, while if α4 > 2 then there ought to be an a3 before
d(1) which there is not. Thus the triple sequence for t is
(0, 0, −γ)(2, 1, 0)(1, 0, γ2)(0, 1, γ2 − 1)(2, 0, γ2 − 2).
If r = 4 then γ2 = 2 and we have
(0, 0, −2)(2, 1, 0)(1, 0, 2)(0, 1, 1)(2, 0, 0)
corresponding to X = E6
0
1
2
3
4
5
E6
with
t = (1 2 3 4 5 0 3 2 1 4 3 2 0 3 4 5)
.
14
The corresponding heap is one of the two minuscule posets for E6 .
1
2
3
0
4
3
5
2
1
4
3
2
0
3
4
5
F (E6 , 5)
If r > 4 then (α5 , β5 , γ5) = (1, 1, γ2 − 3) = (1, 1, 0) which gives γ2 = 3 = γ
and (α6 , β6 , γ6 ) = (0, 0, 3) for maximality, yielding a final sequence
(0, 0, −3)(2, 1, 0)(1, 0, 3)(0, 1, 2)(2, 0, 1)(1, 1, 0)(0, 0, 3)
corresponding to X = E7
0
with
1
2
3
4
5
6
E7
t = (654321032456304532143203456)
The corresponding heap is the unique minuscule poset for E7 , which we
call the swallow. It is perhaps somewhat remarkable that this distributive
lattice is precisely the lattice of order ideals in the either of the minuscule
posets for E6 . This is part of a more general ‘cascading’ phenomenon which
goes back to an observation of Steinberg noted and explained by Proctor in
[P]. The minuscule posets for E6 are themselves lattices of order ideals in the
spin posets for D5 .
15
6
5
4
3
2
0
1
3
2
4
3
5
0
4
6
3
5
2
4
1
3
2
0
3
4
5
F (E7 , 6)
6
This completes the analysis of the case α1 = 2.
Case γ2 = 1:
sequence
We now examine the case r > 3 with γ2 = 1 and triple
(0, 0, −γ)(α1 , 1, 0)(α1 − 1, 0, 1)(α1 − 2, 1, 0).
Then γ = 1 for if γ > 1 the first occurrence of c2 must occur before d(1)
by maximality (since we know c1 occurs before d(1) ), while then the next
occurrence follows at least three c1 ’s, which is impossible. Thus β = γ = 1
and the triple sequence must have the form
(0, 0, −1)(α, 1, 0)(α − 1, 0, 1)(α − 2, 1, 0), · · · , (0, 1, 0) or (0, 0, 1)
depending on the parity of α. Thus X = Dn and
t = 1 2 3 4 · · · n − 1 0 2 3 4 · · · n − 2 1 2 3 0 2 1 · · · 2 3 1 2 0.
These are the same kind of triangular heaps as the example of F (D5 , 4)
pictured earlier.
Finally suppose X has no vertices of degree 3 or more, and t begins with
a vertex d which has two chains emanating from it as shown
• ...
aα a2
a1
d
16
b1
... •
bβ
b2
This is really a special case of the above where now γ = 0, and the same
arguments show that t is of the form
t = d(1) aα1 bβ1 d(2) aα2 bβ2 . . . d(r) aαr bβr .
Note that we have used the assumption that t begins with d. Now by neighbourliness, each αi , βi > 0 for i = 1, · · · , r − 1, and since for αi > 0,
αi+1 = αi − 1 we see that the sequences (α1 , α2 , · · · , αr ), (β1 , β2 , · · · , βr ) are
decreasing incrementally and one must end it zero. It follows that α1 = α1
β1 = β and t is uniquely determined, namely
t = daα bβ daα−1 bβ−1 d · · · damax(0,|α−β|) bmax(0,|α−β|) .
Here for example is the case α = 3, β = 1, corresponding to X A5 .
4
3
2
1
4
3
2
This completes the proof.
5
F (A5 , 2)
3 Connections
The heaps we have constructed are examples of coloured posets, since each
vertex may be considered to be coloured by the corresponding vertex of the
Coxeter graph. If we ignore the labels, these posets are just the irreducible
‘minuscule’ posets defined by Proctor in [4] and shown in figure 2 of Proctor
[5]. As indicated in [4], these posets encapsulate the structure of some of
the most important Bruhat orders on Weyl groups; in fact if an irreducible
Bruhat poset is a lattice then either the Weyl group W is of type G2 or
the poset is isomorphic to the poset induced on the W -orbit of a minuscule
weight with respect to the usual ordering of weights.
These posets play interesting roles in algebraic geometry and Lie theory, including describing the cohomology ring for minuscule flag manifolds
including the Grassmanians. See for example Hiller [2] and Seshadri [7] for
connections with the Schubert calculus of G/P where P is the stabilizer in a
simple Lie group G of a maximal weight space in a minuscule representation.
Minuscule representations have the property that all weights are conjugate under the Weyl group. In this case, the geometry and order structure of
this orbit of weights naturally determines much about the representation. All
of the simply laced simple Lie algebras have minsucule representations with
17
the sole exception of E8 (which is why the latter does not appear in our main
result). For connections with minuscule representations, see Stembridge [9],
Parker and Rohrle [3], and recent work of Donnelly [1].
Some other combinatorial characterizations of minuscule posets appear on
pp 344-345 of [4], including the fact that they constitute all known ‘Gaussian’
posets and that they are exactly the posets of join-irreducibles of the lattice
of weights of minuscule representations of simple Lie algebras. It is also
noted there that minscule posets are strongly Sperner, as well as being rank
unimodal and rank symmetric.
More recently Proctor has shown that the minuscule posets are exactly
the self-dual ”d-complete” posets in [6]. Stembridge has found a new characterization of ”coloured d-complete” posets which consists of (H1) and (H2)
on p8 of [10]. In this language, the posets of this paper are those maximal
amongst those satisfying (H1) and (H2*) which in addition satisfy (H2). Here
(H2*) refers to having at least two elements whose labels are adjacent to i
contained in every open subinterval between two elements labelled i.
References
[1] Donnelly, R. G., Solitary bases of Representations for semisimple Lie
algebras, preprint (2000).
[2] Hiller H., Geometry of Coxeter groups , Research Notes in Mathematics, 54 Pitman(Advanced Publishing Program), Boston, Mass.,
1982.
[3] Parker C. and Rohrle G., Minuscule Representations, preprint (1993).
[4] Proctor R. A., Bruhat Lattices, Plane Partition Generating Functions,
and Minuscule Representations, Europ. J. Combinatorics 5, 331–350
(1984).
[5] Proctor R. A., A Dynkin diagram classification theorem arising from
a combinatorial problem, Advances Math. 62, 103–117 (1986).
[6] Proctor R. A., Dynkin diagram classification of λ-minuscule Bruhat
lattices and of d-complete posets, J. Algebraic Combin. 9, no.1, 61–94
(1999).
[7] Seshadri C. S., Geometry of G/P .I. Theory of standard monomials for
minuscule representations, in C.P. Ramanujam –a tribute, 207–239,
Tata Inst. Fund. Res. Studies in Math., 8, Springer, Berlin, 1978.
[8] Stembridge J. R., On the fully commutative elements of Coxeter
groups, J. Alg. Combins. 5, no.4, 353–385, (1996).
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[9] Stembridge J. R., On minuscule representations, plane partitions and
involutions in complex Lie groups, Duke Math. J., no.2, 469–490
(1994).
[10] Stembridge J. R., Minuscule Elements of Weyl groups, preprint
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[11] Viennot, G. X. Heaps of pieces I: Basic definitions and combinatorial
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[12] Wildberger N. J., A combinatorial construction for simply-laced Lie
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